SOLUTION: How much 20% hydrochloric acid must be mixed with 40 mL of 8% acid to form the 12% acid that is needed in an experiment?

Question 5696: How much 20% hydrochloric acid must be mixed with 40 mL of 8% acid to form the 12% acid that is needed in an experiment?
Answer by prince_abubu(198) (Show Source): You can put this solution on YOUR website!
There is a general formula to use in this kind of problem.

q1*p1 + q2*p2 = (q1 + q2) * p3.

q1 would be the solution's amount that's "already there", AKA, the mixture that you'll be adding some stuff to. p1 would be the concentration (converted to decimal from percent) of q1.

q2 would be the mixture that you're adding. It's more likely to have a different concentration than q1's so we'll name its concentration p2.

Now the right side of the equation will have the quantity q1 + q2. The mixture MUST have the total amount of q1 and q2, but will have a different percentage than either q1 or q2; that's why we're assigning it a different concentration percentage, p3.

In this kind of problem, they will ALWAYS give you ALL values EXCEPT the one you have to find. In this case, they want you to find q2 - the amount of 20% hydrochloric acid that must be added to your original solution.

q1*p1 + q2*p2 = (q1 + q2)* p3
40*0.08 + q2*0.20 = (40 + q2) * 0.12 <------- this would be your setup equation.

3.2 + 0.20 * q2 = 4.8 + 0.12 * q2
0.20 * q2 = 1.6 + 0.12 * q2 <------ subtracted 3.2 from both sides
0.08 * q2 = 1.6 <--------subtracted 0.12 * q2 from both sides
q2 = 20 mL

So, you must add 20 mL of a 20% hydrochloric acid to a 40mL of 8% acid to get a new mixture that is 12% acid.

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